Question

When a conductor is stretched, its length becomes double.
Find out how many times the resistance changes.​

Answer 1

• When the wire is stretched to double the length , the area of cross section gets reduced to half. So when the wire is stretched to 2 times , the resistance multiplies by four times.

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Information about resistance :

• Resistance is directly proportional to the length of the wire.

$\large \underline{ \boxed{ \green{ \bf R \propto l}}}$

• Resistance is inversely proportional to the area of cross section of the wire

$\underline{ \boxed{ \orange{ \bf </strong><strong>R</strong><strong> \propto \frac{1} {</strong><strong>A</strong><strong>} }}}$

Answer 2

Answer: resistance increases by 4 times.

Explanation: in both cases the electricity will be same. And resistance = rho L/A...before stretching the length is 'L ' . After stretching the length is 2L....

Now applying conservation of volume :

V1=V2

L1 * A1 = L2 * A2

L * A1 = 2L * A2

A2 = 1/2 A1 ( area becomes half of the initial)

Now calculating resistance for the stretched wire :

R' = rho L2 / A2

R' = rho * 2L / 1/2* A1

R' = rho * 4 L/ A1

R' = 4 rho* L/ A1

(R = rho * L/ A1)

So, R' = 4 R